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Let $ABC$ be a triangle such that $\angle ACB = \pi/6$ and let $a,b,c$ denote the lengths of the sides opposite to $A,B,C$, respectively. What are the value(s) of x for which $a = x^2 + x + 1, b = x^2-1$ and $c = 2x+1$?

I used law of cosines, then saw that I should apply the identity $a^2 + b^2 = (1/2)[(a+b)^2 + (a-b)^2]$, but I didn't know how to handle the remaining terms. (In other words, I got stuck early in the problem).

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up vote 3 down vote accepted


Applying Law of Cosines, $$\cos \frac\pi6=\frac{a^2+b^2-c^2}{2ab}$$

Now, $$b^2+a^2-c^2=(x^2-1)^2+(x^2+x+1)^2-(2x+1)^2$$ $$=(x^2-1)^2+(x^2-x)(x^2+3x+2)\text{ applying } (a^2-b^2)=(a+b)(a-b)\text{ for the last two terms}$$ $$=(x^2-1)^2+x(x-1)(x+1)(x+2)$$ $$=(x^2-1)\{x^2-1+x(x+2)\}=(x^2-1)(2x^2+x-1)$$

As $x^2-1\ne0$ (which would make $b=0$)

$$\frac{(x^2-1)(2x^2+x-1)}{2(x^2-1)(x^2+x+1)}=\frac{(2x^2+x-1)}{2(x^2+x+1)}=\cos \frac\pi6=\frac{\sqrt3}2$$

Can you complete it now?

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I was able to get that, but I don't know what to do with it – AlanH May 3 '13 at 18:34
@AlanH, form the quadratic equation in $x$ – lab bhattacharjee May 3 '13 at 18:36
by expanding the whole thing on the RHS? – AlanH May 3 '13 at 18:37
@AlanH, edited the answer. The only valid answer seems to be $\frac{\sqrt3+1}2$ – lab bhattacharjee May 3 '13 at 19:05

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